BUG: fillna specify duplicated column that is MultiIndex causes RecursionError

[yuji38kwmt]

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Reproducible Example

import pandas
import numpy

data = [[numpy.nan, 2, 3], [4, numpy.nan, 6], [7, 8, numpy.nan]]
df = pandas.DataFrame(
    data, columns=pandas.MultiIndex.from_tuples([("x", "a"), ("x", "a"), ("y", "b")])
)
df.fillna({("x", "a"): 0})
# RecursionError occurred

Issue Description

I have created a dataframe that contains MultiIndex columns. However, the columns in the dataframe are duplicated.

When I execute the fillna function and specify the duplicated column, a RecursionError occurs.

Expected Behavior

I expected that RecursionError did not occur.

Installed Versions

INSTALLED VERSIONS

commit : 965ceca
python : 3.11.2.final.0
python-bits : 64
OS : Linux
OS-release : 5.15.0-72-generic
Version : #79-Ubuntu SMP Wed Apr 19 08:22:18 UTC 2023
machine : x86_64
processor : x86_64
byteorder : little
LC_ALL : None
LANG : ja_JP.UTF-8
LOCALE : ja_JP.UTF-8

pandas : 2.0.2
numpy : 1.24.2
pytz : 2022.7.1
dateutil : 2.8.2
setuptools : 65.5.0
pip : 23.1.2
Cython : None
pytest : 7.3.1
hypothesis : None
sphinx : None
blosc : None
feather : None
xlsxwriter : None
lxml.etree : 4.9.2
html5lib : 1.1
pymysql : None
psycopg2 : None
jinja2 : 3.1.2
IPython : 8.11.0
pandas_datareader: None
bs4 : None
bottleneck : None
brotli : None
fastparquet : None
fsspec : None
gcsfs : None
matplotlib : None
numba : None
numexpr : None
odfpy : None
openpyxl : None
pandas_gbq : None
pyarrow : 11.0.0
pyreadstat : None
pyxlsb : None
s3fs : None
scipy : None
snappy : None
sqlalchemy : None
tables : None
tabulate : None
xarray : None
xlrd : None
zstandard : None
tzdata : 2023.3
qtpy : None
pyqt5 : None

[JostBrand]

I don't think this is a bug and more of a feature request for fillna to support MultiIndex.:

    def fillna(self, value=None, downcast=None):

        """
        fillna is not implemented for MultiIndex
        """

So the question remains why you get a recursion error and not a NotImplementedError. But as the user documentation points out for duplicate labels:

[...] Other methods, like indexing, can give very surprising results.

I know this solution df['x']['a'].fillna(0) is not really satisfying.